Answer:
a) Probability that exactly 1 fastener is defective, P(X = 1) = 0.144
b) Confidence interval for mean price, ![CI = [4.1058, 4.1202]](https://tex.z-dn.net/?f=CI%20%3D%20%5B4.1058%2C%204.1202%5D)
Step-by-step explanation:
a) Total number of fasteners = 120
Number of defective fasteners = 4
Probability of selecting a defective fastener, p = 4/120
p = 0.033
Probability of selecting an undefective fastener, q = 1 - p
q = 1 - 0.033
q = 0.967
5 fasteners were randomly selected, n =5
Probability that exactly one fastener is defective:

b) Number of gasoline outlets sampled, n = 900
Average gasoline price, 
Standard deviation, 
Confidence Level, CL = 95% = 0.95
Significance level, 

From the standard normal table, 
error margin can be calculated as follows:

The confidence interval will be given as:
![CI = \bar{x} \pm e_{margin} \\CI = 4.113 \pm 0.0072\\CI = [(4.113-0.0072), (4.113+0.0072)]\\CI = [4.1058, 4.1202]](https://tex.z-dn.net/?f=CI%20%3D%20%5Cbar%7Bx%7D%20%5Cpm%20e_%7Bmargin%7D%20%20%5C%5CCI%20%3D%204.113%20%5Cpm%200.0072%5C%5CCI%20%3D%20%5B%284.113-0.0072%29%2C%20%284.113%2B0.0072%29%5D%5C%5CCI%20%3D%20%5B4.1058%2C%204.1202%5D)